Thermodynamics/Gas Law questions

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The discussion revolves around solving two thermodynamics and gas law questions. For the first question, the user successfully calculated the number of moles of gas using the ideal gas law but struggled with finding the mass and density. The mass can be determined by multiplying the number of moles by the molecular weight of the gas. In the second question, the user attempted to find the change in volume of gas as it passed through a jet engine, initially using the wrong method but later corrected their approach using the combined gas law. The final conclusion indicates that the volume remains approximately the same after the changes in pressure and temperature.
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Hi,

I'm stuck on two questions, I gratefull if someone could help me with them, firstly;

(Q1) "A cubical container of volume 0.10m^3 contains a gas at 300K and 1000000Pa. Assuming that the gas is ideal, determine
a) the number of moles of gas present, given that the Universal Gas constant is 8.3 J/K/mol.
b) the mass of gas present given that its relative molecular mass is 352.
c) the density of the gas"

I'm okay I think working out 1a I used the equation n=PV/RT and got 22.22. I can't seem to find how to work out the mass...and 1c I think I'm okay, I guess I just use M/V.

Secondly;

(Q2) "A fixed mass of gas in passing through a jet engine has its pressure increased from 300000 to 1300000Pa, while its temperature rises from 80(deg C) to 1500 (deg C). By what factor does the volume of gas change?"

I worked out an answer to this but it looks wrong, I used P1/T1 = P2/T2 and found the difference between those two to be 116.638, so is that the correct method to work out the change of volume?

Thanks again
 
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(1) Hint: What is the mass of one mole of a material with molecular weight 352?

(2)\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} so \frac{3 \times 10^5 V_1}{353}=\frac{1.3 \times 10^6 V_2}{1773} so V_1=\frac{353 \times 1.3 \times 10^6} {1773 \times 3 \times 10^5} V_2 \approx \frac{5 \times 10^8}{5 \times 10^8} V_2 = V_2
so I get that they should be roughly the same.
 
hey thanks, I get it now
 
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